Question

A pin fin of uniform, cross-sectional area is fabricated of an aluminum alloy (k = 160 W/m-K). The fin diameter is D = 4 mm, and the fin is exposed to convective conditions characterized by h = 220 W/m2-K. It is reported that the fin efficiency is ηf = 0.65. Determine the fin length L and the fin effectiveness εf. Account for tip convection.

Answer 1

Answer:

Given that

D= 4 mm

K = 160 W/m-K

h=h = 220 W/m²-K

ηf = 0.65

We know that

$m=\sqrt{\dfrac{hP}{KA}}$

For circular fin

$m=\sqrt{\dfrac{4h}{KD}}$

$m=\sqrt{\dfrac{4\times 220}{160\times 0.004}}$

m = 37.08

$\eta_f=\dfrac{tanhmL}{mL}$

$0.65=\dfrac{tanh37.08L}{37.08L}$

By solving above equation we get

L= 36.18 mm

The effectiveness for circular fin given as

$\varepsilon =\dfrac{2\ tanhmL}{\sqrt{\dfrac{hD}{K}}}$

$\varepsilon =\dfrac{2\ tanh(37.08\times 0.03618)}{\sqrt{\dfrac{220\times 0.004}{160}}}$

ε = 23.52

Answer 2

Answer:

The length of the fin is 35.2 mm

The effectiveness is 23.5

Solution:

As per the question:

Thermal conductivity, k = $160\ W/m-K$

Diameter of the fin, D = 4 mm = $4\times 10^{- 3}\ m$

Coefficient of convective heat transfer, h = $220\ W/m^{2}-K$

Efficiency of the fin, $\eta_{f} = 0.65$

Now,

To calculate the length if the fin:

$\eta_{f} = \frac{tanh\ mL_{c}}{mL_{c}}$             (1)

where

$L_{c} = L + \frac{D}{4} = (L + 1)\ mm$

Also,

$m = \sqrt{\frac{Ph}{A_{c}k}}$

where

P = Perimeter of the fin = $\pi D$

$A_{c}$ = Convection Area

$A_{c} = \frac{\pi D^{2}}{4}$

Now,

$m = \sqrt{\frac{\pi Dh}{\frac{\pi D^{2}}{4}k}}$

$m = \sqrt{\frac{4h}{Dk}} = \sqrt{\frac{4\times 220}{4\times 10^{- 3}\times 160}} = 37.08\ m^{- 3}$

Now, using eqn (1):

$0.65 = \frac{tanh\ (37.08L_{c})}{37.08L_{c}}$

$tanh\ (37.08L_{c}) = 24.10L_{c}$

Now, by trial-error:

$L_{c} = 0.0362 m = 36.2\ mm$

Also, we know that:

$L_{c} = L + 1$

$36.2 = L + 1$

L = 35.2 mm

Now, the fin effectiveness is given by:

$\epsilon_{f} = \sqrt{\frac{Pk}{hA_{c}}}$

$\epsilon_{f} = \sqrt{\frac{\pi D\times 160}{220\times \frac{\pi D^{2}}{4}}}tanh\ mL_{c}$

$\epsilon_{f} = \sqrt{\frac{4\times 160}{220\times 4\times 10^{- 3}}}tanh\ (37.08\times 0.0362) = 23.5$